Question

The true dip of a normal limb associated with asymmetric folding is 20° and the angle that this limb makes with the axial planar cleavage is 35°. The true dip of the axial planar cleavage, in degree, is.........

Hint:

To calculate the true dip of the axial planar cleavage, use the relationship \( \tan(D_{\text{cleavage}}) = \tan(D_{\text{limb}}) \times \sin(\theta) \), where \( \theta \) is the angle between the limb and cleavage.

Answer 1

Correct Answer: 15

Step 1: List the givens.
The normal limb dips 20 degrees. The angle between this limb and the axial plane cleavage is 35 degrees. We want the cleavage dip.

Step 2: Use the dip relation.
The link between the dips is \[ \tan D_{c} = \tan D_{l} \times \sin\theta, \] where $D_l$ is the limb dip and $\theta$ is the angle between them.

Step 3: Insert the values.
With $\tan 20^\circ \approx 0.364$ and $\sin 35^\circ \approx 0.574$, \[ \tan D_c = 0.364 \times 0.574 \approx 0.209. \]

Step 4: Take the inverse tangent.
So \[ D_c = \tan^{-1}(0.209) \approx 15^\circ. \]

Step 5: State the answer.
The true dip of the axial plane cleavage is about 15 degrees.
\[ \boxed{15^\circ} \]